Use the fact that if $H$ is cyclic and $k|\sharp H$, then $H$ has a unique subgroup of order $k$.
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user10676Feb 5 '12 at 20:18

5

You might want to do it by proving the following two useful statements. 1: If $H$ is characteristic in $N$ and $N$ is normal in $G$ then $H$ is normal in $G$. 2: If $H$ is a subgroup of $G$ and $G$ is cyclic, then $H$ is characteristic in $G$.
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Tobias KildetoftFeb 5 '12 at 20:19

Moreover, for a cyclic group $H$ of order $n$ and every $m \mid n$, there is a unique subgroup of $H$ of order $m$.
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Dylan MorelandFeb 5 '12 at 20:22

I understand what you mean but $G$ is not necessarily a finite group.
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spohreisFeb 5 '12 at 21:27

That's a good point, but there aren't so many infinite cyclic groups!
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Dylan MorelandFeb 6 '12 at 2:21

5 Answers
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Here is a somewhat more general fact which seems useful enough to keep in mind:

If $G$ is a group, $H$ is a normal subgroup of $G$ and $K$ is a characteristic subgroup of $H$, then $K$ is a normal subgroup of $G$.

The proof is almost immediate if you know the definitions: for any $x \in G$, since $H$ is normal in $G$, conjugation by $H$ induces an automorphism $\varphi_x$ of $H$, but not necessarily an "inner" automorphism: i.e., if $x \notin H$, $\varphi_x$ need not be conjugation by any element of $H$. Thus we have assumed that $K$ is just not normal but characteristic as a subgroup of $H$, i.e., stable under all automorphisms of $H$. Done.

As others have pointed out, we also need to see that any subgroup of a cyclic group $H$ is characteristic. Well, any subgroup which is the unique subgroup of its order is characteristic -- this takes care of the case in which $H$ is finite. And any subgroup which is the unique subgroup of its index is characteristic -- this takes care of the case in which $H$ is infinite. (Alternately, if $H \cong (\mathbb{Z},+)$, the only nontrivial automorphism is multiplication by $-1$, which evidently stabilizes all the subgroups $n \mathbb{Z}$.)

Suppose $H = \langle h \rangle$ is normal in $G$ and that $K$ is a subgroup of $H$. Any subgroup of a cyclic group is cyclic, so $K = \langle h^d \rangle$ for some integer $d$.

Let $g \in G$. Since $H$ is normal, $g^{-1}hg = h^i$ for some integer $i$. Then for any integer $k$ you get $g^{-1}(h^d)^kg = (g^{-1}hg)^{dk} = (h^i)^{dk} = (h^d)^{ik}$. This shows that for any $k \in K$, the element $g^{-1}kg$ is in $K$. Therefore $K$ is normal.

I'll give a try. If $H=\langle h \rangle$ , then $H$ is an abelian group and $K$ is a normal subgroup of $H$. Let $d$ the lowest positive integer such that $h^{d}\in K$. Then $K=\langle h^{d} \rangle$ and we have $H/K=\{K,hK,\cdots,h^{d-1}K\}$. Let $g\in G$ and $k=h^{dn}\in K$. Then $gkg^{-1}K=(ghg^{-1})^{dn}K=K$. Thus $gKg^{-1}\subset K$, for all $g\in G$.

You might want to work with the element $gkg^{-1}$ directly instead of the coset $gkg^{-1}K$. That extra $K$ there doesn't really help. You need to use the fact that $ghg^{-1}=...?$ anyway. Remember that the power of a coset of $K$ does not make sense, unless the coset is inside such a group $N$ that we know has $K$ as a normal subgroup :-)
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Jyrki Lahtonen♦Feb 7 '12 at 22:04

@JyrkiLahtonen: Thanks for that. But $gkg^{-1}\in H$ and $H/K$ is a finite abelian group, of order $d$. So $gkg^{-1}K$ has order $d$. Am I correct?
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spohreisFeb 7 '12 at 22:56

Yes. For full marks you need to state that $gkg^{-1}\in H$ (which is, of course, given). But you can then as well argue that because $gkg^{-1}$ has order that is a factor of $n/d$, it has to be in $K$, because $K$ consists precisely of those elements of $H$. I'm just cutting the corner of passing over to $H/K$. +1 for you :-)
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Jyrki Lahtonen♦Feb 8 '12 at 8:09